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प्रश्न
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उत्तर
\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^{- 1} \left( \sin x \right)dx + \int_\frac{\pi}{2}^\pi \sin^{- 1} \left( \sin x \right)dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} xdx + \int_\frac{\pi}{2}^\pi \left( \pi - x \right)dx ..............\left[ \frac{\pi}{2} \leq x \leq \pi \Rightarrow - \pi \leq - x \leq - \frac{\pi}{2} \Rightarrow 0 \leq \pi - x \leq \frac{\pi}{2} \right]\]
\[ = \left.\frac{x^2}{2}\right|_{- \frac{\pi}{2}}^\frac{\pi}{2} + \left.\frac{\left( \pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\frac{\pi}{2}^\pi \]
\[ = \frac{1}{2}\left( \frac{\pi^2}{4} - \frac{\pi^2}{4} \right) - \frac{1}{2}\left( 0 - \frac{\pi^2}{4} \right)\]
\[= 0 + \frac{\pi^2}{8}\]
\[ = \frac{\pi^2}{8}\]
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